Difference between revisions of "2020 AIME II Problems/Problem 1"

Line 8: Line 8:
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]
[[MAA Notice]]
+
{{MAA Notice}}

Revision as of 19:27, 7 June 2020

Problem

Find the number of ordered pairs of positive integers $(m,n)$ such that ${m^2n = 20 ^{20}}$.

Solution

First, we find the prime factorization of $20^{20}$, which is $2^{40}\times5^{20}$. The equation ${m^2n = 20 ^{20}}$ tells us that we want to select a perfect square factor of $20^{20}$, $m^2$. $n$ will be assigned by default. There are $21\times11=231$ ways to select a perfect square factor of $20^{20}$, thus our answer is $\mbox{231}$.

~superagh The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png